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Following the axiom of separation, we define the intersection of a set x as: $$\cap x= \{ y: \forall z (z \in x \Rightarrow y \in z) \}$$ But by the definition of material implication: $$z \in x \Rightarrow y \in z \equiv z \notin x \lor y \in z$$ If we look at the first part of the OR statement: $$z \notin x$$ which would be true if and only if z is not in x.

However, z is arbitrary, and spans over the entirety of the domain of discourse. Hence, any set not being a member of x would satisfy the condition of the intersection class, without regards to y.

Hence all objects in the domain of discourse satisfy y, which would obviously NOT be the Intersection Class.

Note: I am NOT referring to the intuitively meaningful members $\cap x$ which lie within x; these satisfy the condition meaningfully, but only those elements which exist within the Universe of Sets, but not in x.

Surely there must be a flaw in my reasoning, but I can't seem to find it. Could someone please clarify?

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    $\begingroup$ $y$ is the set of objects for which the implication holds for all $z$. Your definition doesn't blow up unless $x= \emptyset$, in which case $\cap x$ is undefined. $\endgroup$ Jul 12, 2020 at 7:21
  • $\begingroup$ Yes we consider all z in the universe of sets (aka domain of discourse.) These include sets which may not necessarily be in the set of interest x. The definition blows up because sets z not in x would automatically satisfy the material implication, regardless of y, and thus true for all y. $\endgroup$
    – Joeseph123
    Jul 12, 2020 at 7:35

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The case $z \not\in x$ doesn't say anything about $y$ at all. Indeed, the only cases we care about are those for which $z \in x$; then we check whether $y \in z$. As Robert Shore wrote, the only case in which we never get to check the case we care about is if $x$ is empty; but then $\bigcap x$ is undefined (some authors define $\bigcap \emptyset = V$).

So, where did your reasoning go wrong? You're implying that if $z \not\in x$ then that would witness that $y \in \bigcap x$. But that's not true; given $x$, the formula $$ z \in x \Rightarrow y \in z $$ must hold for any $z$. Finding a singular $z$ for which the formula holds is not sufficient.

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  • $\begingroup$ If $$z \notin x = True \therefore z \notin x \lor y \in z = True $$ Which is a "tautology in y." I'm using strictly the definitions of predicate calculus, and not the interpretation of what an intersection actually is. $\endgroup$
    – Joeseph123
    Jul 12, 2020 at 9:15
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    $\begingroup$ @RobertShore I'm so sorry, corrected $\endgroup$
    – MacRance
    Jul 12, 2020 at 10:09
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    $\begingroup$ @Joeseph123 What you've written in your comment is correct, however it's not sufficient to deduce that $y$ satisfies the predicate that defines $\bigcap x$. It has to satisfy this $\forall z$, not just some $z$. That's what Robert was alluding to: if you could show that $z \not\in x$ for all $z$ then your conclusion is correct, and any $y \in \bigcap x$. But that holds iff $x$ is empty. Whenever $z \in x$ (i.e. $x$ is not empty), any $y$ that wants to stand a chance to be in $\bigcap x$ must be an element of $z$. $\endgroup$
    – MacRance
    Jul 12, 2020 at 10:13
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It is true that for any $y$, all $z$ that do not fall within $x$ satisfy the condition. But the intersection of $x$ is defined as those $y$ for which every last $z$ satisfies the condition (however it happens to be expressed)—not just those that don't happen to fall within $x$. Therefore the definition does not blow up.

To make this concrete, let $x = \{\{1, 2\}, \{2\}\}$. We see that $2 \in \cap x$, because for all $z \in x, 2 \in z$. However, no other element is in $\cap x$. For example, $1 \not\in \cap x$, because there exists a $z \in x$ such that $1 \not\in z$—namely, $\{2\}$. Equivalently (and this goes to the point of your question), there exists a $z$ such that neither $z \not\in x$ nor $1 \in z$—again, namely, $\{2\}$.

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  • $\begingroup$ My question is not about what happens inside x, but what happens to all objects outside of the set x. $\endgroup$
    – Joeseph123
    Jul 12, 2020 at 9:11
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    $\begingroup$ @Joeseph123 Well, let us look at a representative: $3\notin\bigcap x$ because $\forall z~(z\notin x\vee 3\in z)$ does not hold. There exists a counterexample. When $z$ is instantiated as $\{2\}$, there we evaluate $\{2\}\notin x\vee 3\in\{2\}$ as false. And so forth. $$\neg~\forall z~(z\in x\to y\in z)~~\iff~~\exists z~(z\in x\wedge y\notin z)$$ $\endgroup$ Jul 13, 2020 at 9:45
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    $\begingroup$ No matter how many instances do satisfy its predicate, a universal statement is falsified by the existence of one counter example. So, because, all sets that are not in $x$ must satisfy $z\notin x\vee y\in z$, it is only the sets in $x$ that determine the inclusion of an element in $\bigcap x$. [...unless $x$ is empty...] $\endgroup$ Jul 13, 2020 at 10:07

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